Question

Suppose that 23.21 g of ice at -11.8°C is placed in 61.33 g of water at 93.0°C in a perfectly insulated vessel. Calculate the final temperature. (The molar heat capacity for ice is 37.5 J K^-1 mol^-1 and that for liquid water is 75.3 J K^-1 mol^-1. The molar enthalpy of fusion for ice is 6.01 kJ/mol. You must answer in Kelvin, not °C.)

Answer 1

mass of ice = 23.21 g

we have below equation to be used:

number of mol of ice,

n = mass of ice/molar mass of ice

=(23.21 g)/(18 g/mol)

= 1.289 mol

mass of water = 61.33 g

we have below equation to be used:

number of mol of water,

n = mass of water/molar mass of water

=(61.33 g)/(18 g/mol)

= 3.407 mol

we have:

Latent heat of fusion of ice, Lf = 6010 J/mol

Specific heat capacity of ice, Ci = 37.5 J/mol.K

Specific heat capacity of water, Cw = 75.3 J/mol.K

mol of ice, ni = 1.289 mol

mol of water, nw = 3.407 mol

heat required to take ice from -11.8 oC to 0 oC,

Q1 = ni*Ci*delta Ti

= 1.289*37.5*(0-(-11.8))

= 570.5792 J

heat required to take water from 93.0 oC to 0 oC,

Q2 = nw*Cw*delta Tw

= 3.407*75.3*(93.0-0)

= 23860.4365 J

heat required to melt whole ice,

Q3 = ni*Lf

= 1.289*6010

= 7749.5611 J

Q1 + Q3 = 570.5792+7749.5611=8320.1403<Q2

So whole of ice will melt

Only water will be there in the system

let final temperature be ToC

Q1 + Q3 + heat required by melted ice = heat released by water

570.5792 + 7749.5611 +ni*Cw*deltaT = nw*Cw*delta Tw

570.5792 + 7749.5611 + 1.289*75.3*(T-0) = 3.407*75.3*(93.0-T)

8320.1403 + 97.0952*T = 23860.4365 - 256.5638*T

T = 43.9 oC

= (43.9 + 273.1) K

= 317 K

Answer: 317 K